Reading.

Covering pp. 181-195: the Lagrangian for a system of non relativistic charged particles to zeroth order in : electrostatic energy of a system of charges and .mass renormalization.

A closed system of charged particles.

Consider a closed system of charged particles and imagine there is a frame where they are non-relativistic . In this case we can describe the dynamics using a Lagrangian only for particles. i.e.

If we work t order .

If we try to go to , it’s difficult to only use for particles.

This can be inferred from

because at this order, due to radiation effects, we need to include EM field as dynamical.

Start simple

Start with a system of (non-relativistic) free particles

So in the non-relativistic limit, after dropping the constant term that doesn’t effect the dynamics, our Lagrangian is

The first term is where the second is .

Next include the fact that particles are charged.

Here, working to , where we consider the particles moving so slowly that we have only a Coulomb potential , not .

HERE: these are NOT ‘EXTERNAL’ potentials. They are caused by all the charged particles.

For we have have , but we won’t do this today (tomorrow).

To leading order in , particles only created Coulomb fields and they only “feel” Coulomb fields. Hence to , we have

What’s the , the Coulomb field created by all the particles.

\paragraph{How to find?}

or

where

This is a Poisson equation

(where the time dependence has been suppressed). This has solution

This is the sum of instantaneous Coulomb potentials of all particles at the point of interest. Hence, it appears that should be evaluated in 3.11 at ?

However 3.11 becomes infinite due to contributions of the a-th particle itself. Solution to this is to drop the term, but let’s discuss this first.

Let’s talk about the electrostatic energy of our system of particles.

The first term is zero since for a localized system of charges or higher as .

In the second term

So we have

for

Now substitute 3.11 into 3.14 for

or

The first term is the sum of the electrostatic self energies of all particles. The source of this infinite self energy is in assuming a \underline{point like nature} of the particle. i.e. We modeled the charge using a delta function instead of using a continuous charge distribution.

Recall that if you have a charged sphere of radius

PICTURE: total charge , radius , our electrostatic energy is

Stipulate that rest energy is all of electrostatic origin we get that

This is called the classical radius of the electron, and is of a very small scale .

As a matter of fact the applicability of classical electrodynamics breaks down much sooner than this scale since quantum effects start kicking in.

Our Lagrangian is now

where is the electrostatic potential due to all \underline{other} particles, so we have

and for the system

This is THE Lagrangian for electrodynamics in the non-relativistic case, starting with the relativistic action.

Where we left off.

For a localized charge distribution, we’d arrived at expressions for the scalar and vector potentials far from the point where the charges and currents were localized. This was then used to consider the specific case of a dipole system where one of the charges had a sinusoidal oscillation. The charge positions for the negative and positive charges respectively were

so that our dipole moment is

The scalar potential, to first order in a number of Taylor expansions at our point far from the source, evaluated at the retarded time , was found to be

and our vector potential, also with the same approximations, was

We found that the electric field (neglecting any non-radiation terms that died off as inverse square in the distance) was

Direct computation of the magnetic radiation field

Taking the curl of the vector potential 2.6 for the magnetic field, we’ll neglect the contribution from the since that will be inverse square, and die off too quickly far from the source

which is

Comparing to 2.6, we see that this equals as expected.

An aside: A tidier form for the electric dipole field

We can rewrite the electric field 2.6 in terms of the retarded time dipole

where

Then using the vector identity

we have for the fields

Calculating the energy flux

Our Poynting vector, the energy flux, is

Expanding just the cross terms we have

Note that we’ve utilized to do the cancellations above, and for the final grouping. Since , the direction cosine of the unit radial vector with the z-axis, we have for the direction of the Poynting vector

Our Poynting vector is found to be directed radially outwards, and is

The intensity is constant along the curves

PICTURE: dipole lobes diagram with up along the z axis, and pointing in an arbitrary direction.

FIXME: understand how this lobes picture comes from our result above.

PICTURE: field diagram along spherical north-south great circles, and the electric field along what looks like it is the direction, and along what appear to be the direction, and pointing radially out.

Utilizing the spherical unit vectors to express the field directions.

In class we see the picture showing these spherical unit vector directions. We can see this algebraically as well. Recall that we have for our unit vectors

with the volume element orientation governed by cyclic permutations of

We can now express the direction of the magnetic field in terms of the spherical unit vectors

The direction of the electric field was in the direction of where was directed along the z-axis. This is then

Calculating the power

Integrating over a spherical surface, we can calculate the power

FIXME: remind myself why Power is an appropriate label for this integral.

This is

Averaging over a period kills off the cosine term

and we once again see that higher frequencies radiate more power (i.e. why the sky is blue).

Types of radiation.

We’ve seen now radiation from localized current distributions, and called that electric dipole radiation. There are many other sources of electrodynamic radiation, of which here are a couple.

\begin{itemize}
\item Magnetic dipole radiation.

This will be covered more in more depth in the tutorial. Picture of a positive circulating current given, and a magnetic dipole moment .

This sort of current loop is a source of magnetic dipole radiation.

\item Cyclotron radiation.

This is the label for acceleration induced radiation (at high velocities) by particles moving in a uniform magnetic field.

PICTURE: circular orbit with speed . The particle trajectories are

This problem can be treated as two electric dipoles out of phase by 90 degrees.

Energy momentum conservation.

We’ve defined

(where was defined as the energy flow).

Dimensional analysis arguments and analogy with classical mechanics were used to motivate these definitions, as opposed to starting with the field action to find these as a consequence of a symmetry. We also saw that we had a conservation relationship that had the appearance of a four divergence of a four vector. With

that was

The left had side has the appearance of a Lorentz scalar, since it contracts two four vectors, but the right hand side is the continuum equivalent to the energy term of the Lorentz force law and cannot be a Lorentz scalar. The conclusion has to be that is not a four vector, and it’s natural to assume that these are components of a rank 2 four tensor instead (since we’ve got just one component of a rank 1 four tensor on the RHS). We want to know find out how the EM energy and momentum densities transform.

Classical mechanics reminder.

Recall that in particle mechanics when we had a Lagrangian that had no explicit time dependence

that energy resulted from time translation invariance. We found this by taking the full derivative of the Lagrangian, and employing the EOM for the system to find a conserved quantity

Taking differences we have

and we labeled this conserved quantity the energy

Our approach from the EM field action.

Our EM field action was

The squared field tensor only depends on the fields or its derivatives , and not on the coordinates themselves. This is very similar to the particle action with no explicit time dependence

For the particle case we obtained our conservation relationship by taking time derivatives of the Lagrangian. These are very similar with the action having no explicit dependence on space or time, only on the field, so what will we get if we take the coordinate partials of the EM Lagrangian density?

We will chew on this tomorrow and calculate

in full gory details. We will find that instead of finding a single conserved quantity , we instead find a quantity that only changes through escape from the boundary of a surface.

This difference in power shows the second order moment dependence, in the approximations.

FIXME: go back and review the “third year” content and see where the magnetic dipole moment came from. That’s the key to this argument, since we need to see how this ends up equivalent to a pair of charges in the electric field case.

Midterm solution discussion.

In the last part of the tutorial, the bonus question from the tutorial was covered. This was to determine the Yukawa potential from the differential equation that we found in the earlier part of the problem.

I took a couple notes about this on paper, but don’t intend to write them up. Everything proceeded exactly as I would have expected them to for solving the problem (I barely finished the midterm as is, so I didn’t have a chance to try it). Take Fourier transforms and then evaluate the inverse Fourier integral. This is exactly what we can do for the Coulomb potential, but actually easier since we don’t have to introduce anything to offset the poles (and we recover the Coulomb potential in the case).

Multipole expansion of the fields.

This integral is over the region of space where the sources are non-vanishing, but this region is limited. The value , so we can expand the denominator in multipole expansion

Neglecting all but the first order term in the expansion we have

Similarly, for the retarded time we have

We can now do a first order Taylor expansion of the current about the retarded time

To elucidate the physics, imagine that time dependence of the source is periodic with angular frequency . For example:

Here we have

So, for the magnitude of the second term we have

Requiring second term much less than the first term means

But recall

so for our Taylor expansion to be valid we have the following constraints on the angular velocity and the position vectors for our charge and measurement position

This is a physical requirement size of the wavelength of the emitter (if the wavelength doesn’t meet this requirement, this expansion does not work). The connection to the wavelength can be observed by noting that we have

Putting the pieces together. Potentials at a distance.

\paragraph{Moral:} We’ll utilize two expansions (we need two small parameters)

\begin{enumerate}
\item
\item
\end{enumerate}

Plugging into our current

The first term is the total charge evaluated at the retarded time. In the second term (and in the third, where it’s derivative is taken) we have

which is the dipole moment evaluated at the retarded time . In the last term we can pull out the time derivative (because we are integrating over )

For the spatial components of the current lets just keep the first term

There’s two tricks used here. One was writing the unit vector . The other was use of the continuity equation . This first trick was mentioned as one of the few tricks of physics that will often be repeated since there aren’t many good ones.

With the first term vanishing on the boundary (since is localized), and pulling the time derivatives out of the integral, we can summarize the dipole potentials as

Example: Electric dipole radiation

PICTURE: two closely separated oppositely charges, wiggling along the line connecting them (on the z-axis). at rest, while oscillates.

Since we’ve put the charge at the origin, it has no contribution to the dipole moment, and we have

Thus

so with , and in the dipole dot product, we have

These hold provided and . Recall that , which has dimensions of velocity.

FIXME: think through and justify .

Observe that so this is a requirement that our charged positive particle is moving with .

Now we’ll take derivatives. The first term of the scalar potential will be ignored since the is non-radiative.

We’ve used , and , and .

So,

The power is proportional to . Higher frequency radiation has more power : this is why the sky is blue! It all comes from the fact that the electric field is proportional to the squared acceleration ().

Problem 1. Energy, momentum, etc., of EM waves.

Statement

\begin{enumerate}
\item Calculate the energy density, energy flux, and momentum density of a plane monochromatic linearly polarized electromagnetic wave.
\item Calculate the values of these quantities averaged over a period.
\item Imagine that a plane monochromatic linearly polarized wave incident on a surface (let the angle between the wave vector and the normal to the surface be ) is completely reflected. Find the pressure that the EM wave exerts on the surface.
\item To plug in some numbers, note that the intensity of sunlight hitting the Earth is about ( the intensity is the average power per unit area transported by the wave). If sunlight strikes a perfect absorber, what is the pressure exerted? What if it strikes a perfect reflector? What fraction of the atmospheric pressure does this amount to?
\end{enumerate}

Solution

Part 1. Energy and momentum density.

Because it doesn’t add too much complexity, I’m going to calculate these using the more general elliptically polarized wave solutions. Our vector potential (in the Coulomb gauge , ) has the form

The elliptical polarization case only differs from the linear by allowing to be complex, rather than purely real or purely imaginary. Observe that the Coulomb gauge condition implies

a fact that will kill of terms in a number of places in the following manipulations.

Also observe that for this to be a solution to the wave equation operator

the frequency and wave vector must be related by the condition

For the time and spatial phase let’s write

In the Coulomb gauge, our electric and magnetic fields are

Similar to \S 48 of the text [1], let’s split into a phase and perpendicular vector components so that

where has a real square

This allows a split into two perpendicular real vectors

where since is real.

Our electric and magnetic fields are now reduced to

or explicitly in terms of and

The special case of interest for this problem, since it only strictly asked for linear polarization, is where and one of or is zero (i.e. is strictly real or strictly imaginary). The case with strictly real, as done in class, is

Now lets calculate the energy density and Poynting vectors. We’ll need a few intermediate results.

and

Let’s use arrowed vectors for the phasor parts

where we can recover our vector quantities by taking real parts , . Our energy density in terms of these phasors is then

This is

Note that , and (since ). Also , so we have

Now, for the Poynting vector. We have

This is

Reducing the terms we get , and , leaving

Now, the text in \S 47 defines the energy flux as the Poynting vector, and the momentum density as , so we just divide 1.22 by for the momentum density and we are done. For the linearly polarized case (all that was actually asked for, but less cool to calculate), where is real, we have

Part 2. Averaged.

We want to average over one period, the time such that , so the average is

It is clear that this will just kill off the sinusoidal terms, leaving

Part 3. Pressure.

The magnitude of the momentum of light is related to its energy by

and can thus loosely identify the magnitude of the force as

With pressure as the force per area, we could identify

as the instantaneous (directed) pressure on a surface. What is that for linearly polarized light? We have from above for the linear polarized case (where )

If we look at the magnitude of the average pressure from the radiation, we have

Part 4. Sunlight.

With atmospheric pressure at , and the pressure from the light at , we have roughly of pressure from the sunlight being only of the total atmospheric pressure. Wow. Very tiny!

Would it make any difference if the surface is a perfect absorber or a reflector? Consider a ball hitting a wall. If it manages to embed itself in the wall, the wall will have to move a bit to conserve momentum. However, if the ball bounces off twice the momentum has been transferred to the wall. The numbers above would be for perfect absorbtion, so double them for a perfect reflector.

Problem 2. Spherical EM waves.

Statement

Suppose you are given:

where and is the unit vector in the -direction. This is a simple example of a spherical wave.

\begin{enumerate}
\item Show that obeys all four Maxwell equations in vacuum and find the associated magnetic field.
\item Calculate the Poynting vector. Average over a full cycle to get the intensity vector . Where does it point to? How does it depend on ?
\item Integrate the intensity vector flux through a spherical surface centered at the origin to find the total power radiated.
\end{enumerate}

Solution

Part 1. Maxwell equation verification and magnetic field.

Our vacuum Maxwell equations to verify are

We’ll also need the spherical polar forms of the divergence and curl operators, as found in \S 1.4 of [2]

We can start by verifying the divergence equation for the electric field. Observe that our electric field has only an component, so our divergence is

We have a zero divergence since the component has no dependence (whereas itself does since the unit vector ).

All of the rest of Maxwell’s equations require so we’ll have to first calculate that before progressing further.

A aside on approaches attempted to find

I tried two approaches without success to calculate . First I hoped that I could just integrate to obtain and then take the curl. Doing so gave me a result that had . I hunted for an algebraic error that would account for this, but could not find one.

The second approach that I tried, also without success, was to simply take the cross product . This worked in the monochromatic plane wave case where we had

since one can easily show that . Again, I ended up with a result for that did not have a zero divergence.

Finding with a more systematic approach.

Following [3] \S 16.2, let’s try a phasor approach, assuming that all the solutions, whatever they are, have all the time dependence in a term.

Let’s write our fields as

Substitution back into Maxwell’s equations thus requires equality in the real parts of

With we can now directly compute the magnetic field phasor

The electric field of this problem can be put into phasor form by noting

which allows for reading off the phasor part directly

Now we can compute the magnetic field phasor . Since we have only a component in our field, the curl will have just and components. This is reasonable since we expect it to be perpendicular to .

Chugging through all the algebra we have

so our magnetic phasor is

Multiplying by and taking real parts gives us the messy magnetic field expression

Since this was constructed directly from , this implicitly verifies one more of Maxwell’s equations, leaving only , and . Neither of these looks particularly fun to verify, however, we can take a small shortcut and use the phasors to verify without the explicit time dependence.

From 2.54 we have for the divergence

Let’s also verify the last of Maxwell’s equations in phasor form. The time dependence is knocked out, and we want to see that taking the curl of the magnetic phasor returns us (scaled) the electric phasor. That is

With only and components in the magnetic phasor we have

Immediately, we see that with no explicit dependence in the coordinates, we have no nor terms in the curl, which is good. Our curl is now just

What we expect is which is

FIXME: Somewhere I must have made a sign error, because these aren’t matching! Have an extra term and the wrong sign on the term.

Part 2. Poynting and intensity.

Our Poynting vector is

which we could calculate from 2.34, and 2.55. However, that looks like it’s going to be a mess to multiply out. Let’s use instead the trick from \S 48 of the course text [1], and work with the complex quantities directly, noting that we have

Now we can do the Poynting calculation using the simpler relations 2.52, 2.54.

Let’s also write

where

So our Poynting vector is

Note that our unit vector basis was rotated from , so we have

and plug this into our Poynting expression

Now we have to multiply out our terms. We have

Since this has no real part, there is no average contribution to in the direction. What do we have for the time dependent part

This is non zero, so we have a time dependent contribution that averages out. Moving on

This is non-zero, so the steady state Poynting vector is in the outwards radial direction. The last piece is

Assembling all the results we have

We can read off the intensity directly

Part 3. Find the power.

Through a surface of radius , integration of the intensity vector 2.68 is

Our average power through the surface is therefore

Notes on grading of my solution.

Problem 2 above was the graded portion.

FIXME1: I lost a mark in the spot I expected, where I failed to verify one of the Maxwell equations. I’ll still need to figure out what got messed up there.

What occured to me later, also mentioned in the grading of the solution was that Maxwell’s equations in the space-time domain could have been used to solve for instead of all the momentum space logic (which simplified some things, but probably complicated others).

FIXME2: I lost a mark on 2.68 with a big beside it. I’ll have to read the graded solution to see why.

FIXME3: Lost a mark for the final average power result 2.69. Again, I’ll have to go back and figure out why.

Motivation

While this isn’t part of the course, the topic of waveguides is one of so many applications that it is worth a mention, and that will be done in this tutorial.

We will setup our system with a waveguide (conducting surface that confines the radiation) oriented in the direction. The shape can be arbitrary

PICTURE: cross section of wacky shape.

At the surface of a conductor.

At the surface of the conductor (I presume this means the interior surface where there is no charge or current enclosed) we have

If we are talking about the exterior surface, do we need to make any other assumptions (perfect conductors, or constant potentials)?

Wave equations.

For electric and magnetic fields in vacuum, we can show easily that these, like the potentials, separately satisfy the wave equation

Taking curls of the Maxwell curl equations above we have

but we have for vector

which gives us a pair of wave equations

We still have the original constraints of Maxwell’s equations to deal with, but we are free now to pick the complex exponentials as fundamental solutions, as our starting point

With and this is

For the vacuum case, with monochromatic light, we treated the amplitudes as constants. Let’s see what happens if we relax this assumption, and allow for spatial dependence (but no time dependence) of and . For the LHS of the electric field curl equation we have

Similarly for the divergence we have

This provides constraints on the amplitudes

Applying the wave equation operator to our phasor we get

So the momentum space equivalents of the wave equations are

Observe that if , then these amplitudes are harmonic functions (solutions to the Laplacian equation). However, it doesn’t appear that we require such a light like relation for the four vector .

Back to the tutorial notes.

In class we went straight to an assumed solution of the form

where . Our Laplacian was also written as the sum of components in the propagation and perpendicular directions

With no dependence in the amplitudes we have

Separation into components.

It was left as an exercise to separate out our Maxwell equations, so that our field components and in the propagation direction, and components in the perpendicular direction are separated

We can do something similar for . This allows for a split of 1.14 into and perpendicular components

So we see that once we have a solution for and (by solving the wave equation above for those components), the components for the fields in terms of those components can be found. Alternately, if one solves for the perpendicular components of the fields, these propagation components are available immediately with only differentiation.

In the case where the perpendicular components are taken as given

we can express the remaining ones strictly in terms of the perpendicular fields

Is it at all helpful to expand the double cross products?

This gives us

but that doesn’t seem particularly useful for completely solving the system? It appears fairly messy to try to solve for and given the propagation direction fields. I wonder if there is a simplification available that I am missing?

Solving the momentum space wave equations.

Back to the class notes. We proceeded to solve for and from the wave equations by separation of variables. We wish to solve equations of the form

Write , so that we have

One solution is sinusoidal

The example in the tutorial now switched to a rectangular waveguide, still oriented with the propagation direction down the z-axis, but with lengths and along the and axis respectively.

Writing , and , we have

We were also provided with some definitions

\begin{definition}TE (Transverse Electric)

.
\end{definition}
\begin{definition}
TM (Transverse Magnetic)

.
\end{definition}
\begin{definition}
TM (Transverse Electromagnetic)

.
\end{definition}

\begin{claim}TEM do not existing in a hollow waveguide.
\end{claim}

Why: I had in my notes

and then

In retrospect I fail to see how these are connected? What happened to the term in the curl equation above?

It was argued that we have on the boundary.

So for the TE case, where , we have from the separation of variables argument

No sines because

The quantity

is called the mode. Note that since an ampere loop requires since there is no current.

Writing

Since we have purely imaginary, and the term

represents the die off.

is the smallest.

Note that the convention is that the in is the bigger of the two indexes, so .

The phase velocity

However, energy is transmitted with the group velocity, the ratio of the Poynting vector and energy density

(This can be shown).

Since

We see that the energy is transmitted at less than the speed of light as expected.

Final remarks.

I’d started converting my handwritten scrawl for this tutorial into an attempt at working through these ideas with enough detail that they self contained, but gave up part way. This appears to me to be too big of a sub-discipline to give it justice in one hours class. As is, it is enough to at least get an concept of some of the ideas involved. I think were I to learn this for real, I’d need a good text as a reference (or the time to attempt to blunder through the ideas in much much more detail).

Reading.

Covering lecture notes pp. 136-146: continued reminder of electrostatic Greens function (136); the retarded Greens function of the d’Alembert operator: derivation and properties (137-140); the solution of the d’Alembert equation with a source: retarded potentials (141-142)

Solving the forced wave equation.

See the notes for a complex variables and Fourier transform method of deriving the Green’s function. In class, we’ll just pull it out of a magic hat. We wish to solve

(with a gauge choice).

Our Green’s method utilizes

If we know such a function, our solution is simple to obtain

Proof:

Claim:

This is the retarded Green’s function of the operator , where

Proof of the d’Alembertian Green’s function

Our Prof is excellent at motivating any results that he pulls out of magic hats. He’s said that he’s included a derivation using Fourier transforms and tricky contour integration arguments in the class notes for anybody who is interested (and for those who also know how to do contour integration). For those who don’t know contour integration yet (some people are taking it concurrently), one can actually prove this by simply applying the wave equation operator to this function. This treats the delta function as a normal function that one can take the derivatives of, something that can be well defined in the context of generalized functions. Chugging ahead with this approach we have

This starts things off and now things get a bit hairy. It’s helpful to consider a chain rule expansion of the Laplacian

In vector form this is

Applying this to the Laplacian portion of 2.6 we have

Here we make the identification

This could be considered a given from our knowledge of electrostatics, but it’s not too much work to just do so.

An aside. Proving the Laplacian Green’s function.

If is a Green’s function for the Laplacian, then the Laplacian of the convolution of this with a test function should recover that test function

We can directly evaluate the LHS of this equation, following the approach in [2]. First note that the Laplacian can be pulled into the integral and operates only on the presumed Green’s function. For that operation we have

It will be helpful to compute the gradient of various powers of

In particular we have, when , this gives us

For the Laplacian of , at the points where this is well defined we have

So we have a zero. This means that the Laplacian operation

can only have a value in a neighborhood of point . Writing we have

Observing that we can put this in a form that allows for use of Stokes theorem so that we can convert this to a surface integral

where we use as the outwards normal for a sphere centered at of radius . This integral is just , so we have

The convolution of with produces , allowing an identification of this function with a delta function, since the two have the same operational effect

Returning to the d’Alembertian Green’s function.

We need two additional computations to finish the job. The first is the gradient of the delta function

Consider . This is

so we have

The Laplacian is similar

so we have

With , we’ll need the Laplacian of this vector magnitude

So that we have

Now we have all the bits and pieces of 2.8 ready to assemble

Since we also have

The terms cancel out in the d’Alembertian, leaving just

Noting that the spatial delta function is non-zero only when , which means in this product, and we finally have

We write

Elaborating on the wave equation Green’s function

The Green’s function 2.26 is a distribution that is non-zero only on the future lightcone. Observe that for we have

We say that is supported only on the future light cone. At , only the contributions for matter. Note that in the “old days”, Green’s functions used to be called influence functions, a name that works particularly well in this case. We have other Green’s functions for the d’Alembertian. The one above is called the retarded Green’s functions and we also have an advanced Green’s function. Writing for advanced and for retarded these are

There are also causal and non-causal variations that won’t be of interest for this course.

This arms us now to solve any problem in the Lorentz gauge

The additional EM waves are the possible contributions from the homogeneous equation.

Since is non-zero only when , the non-homogeneous parts of 3.28 reduce to

Our potentials at time and spatial position are completely specified in terms of the sums of the currents acting at the retarded time . The field can only depend on the charge and current distribution in the past. Specifically, it can only depend on the charge and current distribution on the past light cone of the spacetime point at which we measure the field.

Example of the Green’s function. Consider a charged particle moving on a worldline

( for classical)

For this particle

PICTURE: light cones, and curved worldline. Pick an arbitrary point , and draw the past light cone, looking at where this intersects with the trajectory

For the arbitrary point we see that this point and the retarded time obey the relation

Motivation.

Long solenoid of radius , n turns per unit length, current . Coaxial with with solenoid are two long cylindrical shells of length and of , and respectively, where .

When current is gradually reduced what happens?

The initial fields.

Initial Magnetic field.

For the initial static conditions where we have only a (constant) magnetic field, the Maxwell-Ampere equation takes the form

\paragraph{On the name of this equation}. In notes from one of the lectures I had this called Maxwell-Faraday equation, despite the fact that this isn’t the one that Maxwell made his displacement current addition. Did the Professor call it that, or was this my addition? In [2] Faraday’s law is also called the Maxwell-Faraday equation. [1] calls this the Ampere-Maxwell equation, which makes more sense.

Put into integral form by integrating over an open surface we have

The current density passing through the surface is defined as the enclosed current, circulating around the bounding loop

so by Stokes Theorem we write

Now consider separately the regions inside and outside the cylinder. Inside we have

Outside of the cylinder we have the equivalent of loops, each with current , so we have

Our magnetic field is constant while is constant, and in vector form this is

Torque and angular momentum induced by the fields.

Our torque on the outer cylinder (radius ) that is induced by changing the current is

This provides the induced angular momentum on the outer cylinder

This is the angular momentum of induced by changing the current or changing the magnetic field.

On the inner cylinder we have

So our induced angular momentum on the inner cylinder is

The total angular momentum in the system has to be conserved, and we must have

At the end of the tutorial, this sum was equated with the field angular momentum density , but this has different dimensions. In fact, observe that the volume in which this angular momentum density is non-zero is the difference between the volume of the solenoid and the inner cylinder

so if we are to integrate the angular momentum density 1.17 over this region we have

which does match with the sum of the mechanical angular momentum densities 1.24 as expected.

Reading.

Covering lecture notes pp. 128-135: energy flux and momentum density of the EM wave (128-129); radiation pressure, its discovery and significance in physics (130-131); EM fields of moving charges: setting up the wave equation with a source (132-133); the convenience of Lorentz gauge in the study of radiation (134); reminder on Greens functions from electrostatics (135) [Tuesday, Mar. 8]

Review. Energy density and Poynting vector.

Last time we showed that Maxwell’s equations imply

In the lecture, Professor Poppitz said he was free here to use a full time derivative. When asked why, it was because he was considering and here to be functions of time only, since they were measured at a fixed point in space. This is really the same thing as using a time partial, so in these notes I’ll just be explicit and stick to using partials.

Any change in the energy must either due to currents, or energy escaping through the surface.

The energy flux of the EM field: this is the energy flowing through in unit time ().

How about electromagnetic waves?

In a plane wave moving in direction .

PICTURE: , , .

So, since .

for a plane wave is the amount of energy through unit area perpendicular to in unit time.

So we wee that is indeed rightly called “the momentum density” of the EM field.

We will later find that and are components of a rank-2 four tensor

where is the stress tensor. We will get to all this in more detail later.

For EM wave we have

(this is the energy flux)

(the momentum density of the wave).

(recall for massless particles.

EM waves carry energy and momentum so when absorbed or reflected these are transferred to bodies.

Kepler speculated that this was the fact because he had observed that the tails of the comets were being pushed by the sunlight, since the tails faced away from the sun.

Maxwell also suggested that light would extort a force (presumably he wrote down the “Maxwell stress tensor” that is named after him).

This was actually measured later in 1901, by Peter Lebedev (Russia).

PICTURE: pole with flags in vacuum jar. Black (absorber) on one side, and Silver (reflector) on the other. Between the two of these, momentum conservation will introduce rotation (in the direction of the silver).

This is actually a tricky experiment and requires the vacuum, since the black surface warms up, and heats up the nearby gas molecules, which causes a rotation in the opposite direction due to just these thermal effects.

Another example (a factor) that prevents star collapse under gravitation is the radiation pressure of the light.

Moving on. Solving Maxwell’s equation

Our equations are

where we assume that is a given. Our task is to find , the fields.

Proceed by finding . First, as usual when . The Bianchi identity is satisfied so we focus on the current equation.

In terms of potentials

or

We want to work in the Lorentz gauge . This is justified by the simplicity of the remaining problem

Write

where

This is the d’Alembert operator (“d’Alembertian”).

Our equation is

(in the Lorentz gauge)

If we learn how to solve (**), then we’ve learned all.

Method: Green’s function’s

In electrostatics where , only, we have

Solution

PICTURE:

(a small box)

acting through distance , acting at point . With , we have

Also since is deemed a linear operator, we have , we find

We end up finding that

thus solving the problem. We wish next to do this for the Maxwell equation 4.20.

The Green’s function method is effective, but I can’t help but consider it somewhat of a cheat, since one has to through higher powers know what the Green’s function is. In the electrostatics case, at least we can work from the potential function and take it’s Laplacian to find that this is equivalent (thus implictly solving for the Green’s function at the same time). It will be interesting to see how we do this for the forced d’Alembertian equation.